Question

A small, 300 gg cart is moving at 1.30 m/sm/s on a frictionless track when it collides with a larger, 4.00 kgkg cart at rest. After the collision, the small cart recoils at 0.860 m/sm/s . Part A What is the speed of the large cart after the collision

Answer 1

The speed of the Big Dipper after the collision is 0.091 m/s.

Step-by-step explanation:

In a collision, the total momentum before the collision is equal to the total momentum after the collision. We can apply this principle to solve this problem.

First calculate the momentum of the car before the collision:

Small car in front = mass of the small car x speed of the small car in front

Small Momentum Car Front = 0.300 kg x 1.30 m/sec

MomentumSmall car, front = 0.39 kg*m/s

Since the big dipper was at rest before the collision, its momentum is zero. After the collision, the two cars stay together and move at the same final speed. We can determine this speed using the momentum conservation equation:

Momentum small wagon, before + Momentum big wagon, before = Momentum small wagon + big wagon, after

0.39 kg*m/s + 0 kg*m/s = (0.300 kg + 4.0 kg) x next speed

Velocity = 0.39 kg*m/s / 4.30 kg = 0.091 m/s

Therefore, the speed of the big dipper after the collision is 0.091 m/s.

Answer 2

v₂f = 0.16 m/s (to the right)

Step-by-step explanation:

• Assuming no external forces acting during the collision, as stated by Newton's 2nd law, total momentum must be conserved.
• The initial momentum is due to the small cart only, because the big one is at rest.
• Assuming that the small cart is going to the right and taking this direction as the positive one, we can write the following expression for the initial momentum p₀ :

`p_(o) = m_(1) * v_(o1) = 0.3 kg * 1.3 m/s (1)`

• The final momentum can be written as follows:

`p_(f) = m_(1) * v_(1f) + m_(2) * v_(2f) = 0.3 kg* (-0.86m/s) + 4.00 kg* v_(2f) (2)`

As p₀ = pf, from (1) and (2), we can solve for v2f, as follows:

`v_(2f) = (0.3kg* (1.3 + 0.86) m/s)/(4.00kg) = (0.3kg*2.16m/s)/(4.00kg) = 0.16 m/s`

• As the sign of v2f is positive, we conclude that it starts to move in the same direction that m1 was originally going (to the right), whilst m1 recoils, which means that after the collision, it bounces back from the larger mass.